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Math 157 Honors Calculus I Note 1, 9/15/2016 Let Q and I denote the set of rational and irrational numbers, respectively. Thus x ∈ Q means that the number x can be written as the fraction m/n for some integers m, n with n 6= 0, while x ∈ I means that no such representation of x as a fraction is possible. Claim 1. Suppose x and y are both rational. Then x + y and xy are rational. If y 6= 0, then x/y is rational. This is straightforward. By the assumption we can write x = m/n and y = m0 /n0 for some integers m, m0 , n, n0 . Then mn0 + m0 n mm0 and xy = . nn0 nn0 Since mn0 + m0 n, mm0 and nn0 are all integers, it follows that x + y ∈ Q and xy ∈ Q. If y 6= 0, we have m0 6= 0, so mn0 x = . y nm0 Since mn0 and nm0 are integers, we conclude that x/y ∈ Q. x+y = Claim 2. Suppose x is rational and y is irrational. Then (i) x + y is irrational. (ii) xy is irrational unless x = 0 (in which case xy = 0 is rational). (iii) x/y is irrational unless x = 0 (in which case x/y = 0 is rational). To see (i), suppose x + y were rational. Then by Claim 1, − x = (−1) x ∈ Q and so y = − x + ( x + y) ∈ Q, which would contradict our assumption. Hence x + y ∈ I. The argument for (ii) is similar. Let x 6= 0 and suppose xy were rational. Then by Claim 1, 1/x ∈ Q and so y = (1/x )( xy) ∈ Q, which again would contradict our assumption. Hence xy ∈ I. Finally for (iii), suppose x 6= 0 and x/y where rational. Then by what we just proved in part (ii), x = ( x/y)y ∈ I which would be a contradiction. Hence x/y ∈ I. Claim 3. Suppose x and y are irrational. Then x + y and xy may be rational or irrational. √ To see this, let x = −y = 2. Then x ∈ I and y ∈ I, but x + y = 0 ∈ Q √ √ and xy = 2 ∈ Q. On the other √ hand, let x = 2 and y = √3. Then √ x ∈ I and y ∈ I, but this time xy = 6√∈ I. √The sum x + y = 2 + 3 is also irrational. To see √ this, suppose z = 2 + 3 ∈2 Q. Then√by Claim 1 we would 2 have z = 5 + 2 6 ∈ Q. But √ which in turn √ this would imply z − 5 = 2 6 ∈ Q, 2 would show (z − 5)/2 = 6 ∈ Q, which is a contradiction since 6 is irrational. Claim 3 shows that sums and products of irrational numbers could be rational or irrational. We can prove a sharper statement: Claim 4. Every real number x is the sum of two irrational numbers. If x 6= 0, then x is the product of two irrational numbers. √ √ √ To see this, first suppose x ∈ Q. Then, x − 2 ∈ I (why?) so x =√ 2 + ( x − 2) is the sum √ of two √ irrationals. On the other hand, if x 6= 0, then x/ 2 ∈ I (why?) so x = 2 ( x/ 2) is the product of two irrationals. Now suppose x ∈ I. Then x/2 ∈ I (why?) sop x = ( x/2) + ( x/2) is the sum of p two pirrationals. On the other hand, | x | ∈ I so | x | ∈ I (why?). Hence | x | = | x | | x | is the product of two irrationals. Since x is either | x | or −| x |, it easily follows that x itself must be the product of two irrational numbers.